2
$\begingroup$

Evaluate $$P=\int_{0}^{\frac{\pi}{4}} \ln(\sec x)dx$$

My try: I tried using its complimentary integral:

Let $$Q=\int_{0}^{\frac{\pi}{4}} \ln(\csc x)dx$$

Adding both we get:

$$P+Q=\int_{0}^{\frac{\pi}{4}}\ln(\sec x\csc x)dx$$ $\implies$

$$2P+2Q=\int_{0}^{\frac{\pi}{4}}\ln(\sec^2 x\csc^2 x)dx=\int_{0}^{\frac{\pi}{4}}\ln\left(\frac{4}{4\sin^2 x\cos^2 x}\right)dx$$ $\implies$

$$2P+2Q=\frac{\pi}{4}\ln 4-\int_{0}^{\frac{\pi}{4}}\ln\left(\sin^2 2x\right)dx$$

$$2P+2Q=\frac{\pi}{2}\ln 2-2 \int_{0}^{\frac{\pi}{4}}\ln(\sin 2x)dx$$

Using the substitution $2x=t$ we get

$$2P+2Q=\frac{\pi}{2}\ln 2- \int_{0}^{\frac{\pi}{2}}\ln(\sin t)dt$$

Using the formula:

$$\int_{0}^{\frac{\pi}{2}}\ln(\sin t)dt=\frac{-\pi}{2}\ln 2$$ we get

$$2P+2Q=\pi \ln 2$$

$$P+Q=\frac{\pi}{2}\ln 2$$

Is there any way to find $P-Q$

$\endgroup$
2
$\begingroup$

In fact \begin{eqnarray*} P-Q&=&\int_0^{\frac{\pi}{4}}\ln(\tan t)dt\\ &=&\int_0^1\frac{\ln u}{1+u^2}du\\ &=&\int_0^1\ln u\sum_{n=0}^\infty(-1)^nu^{2n}du\\ &=&\sum_{n=0}^\infty(-1)^n\int_0^1u^{2n}\ln udu\\ &=&-\sum_{n=0}^\infty(-1)^n\frac{1}{(2n+1)^2}\\ &=&-C \end{eqnarray*} where $C$ is the Catalan constant.

$\endgroup$
2
$\begingroup$

Through the properties of the logarithm, the integral can be rewritten as$$\mathfrak{I}=-\int\limits_0^{\pi/4}\mathrm dx\,\log\cos x$$Now use the Fourier expansion for $\log\cos x$ which is

$$\log\cos x=\sum\limits_{n\geq1}\frac {(-1)^{n-1}\cos 2nx}{n}-\log 2$$

Now integrate each time termwise to get that$$\begin{align*}\mathfrak{I} & =\frac {\pi}4\log 2+\sum\limits_{n\geq1}\frac {(-1)^n}{n}\int\limits_0^{\pi/4}\mathrm dx\,\cos 2nx\\ & =\frac {\pi}4\log 2+\frac 12\sum\limits_{n\geq1}\frac {(-1)^n}{n^2}\sin\left(\frac {\pi n}2\right)\\ & =\frac {\pi}4\log 2-\frac 12G\end{align*}$$where $G$ is Catalan’s constant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.